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Infinite sheets at $x=0$ and $y=0$ with surface charge density $+\sigma$ and $-\sigma$ have piecewise uniform electric fields. Gauss's law and superposition give the displayed result outside the sheets. The electric field lines bisect each quadrant and connect the positive sheet to the negative one. Its magnitude is $\sigma /(\sqrt{2}{ϵ}_0)$.

The nonuniform electric field near the edges of finite electrodes, where electric field lines bend away from an ideal uniform or one-dimensional configuration. A parallel-plate capacitor with separation much smaller than its lateral size has nearly uniform interior field, but its edge field contributes a correction to capacitance.

Use a cross-section normal to the sheets' line of intersection: the positive sheet is $x=0$, and the negative sheet is $y=0$. A Gaussian pillbox gives a single infinite sheet's electric field magnitude $\sigma /(2{ϵ}_0)$ on either side. Superposition consequently gives, away from the sheets,The electric field points away from the positively charged sheet and towards the negatively charged sheet, bisecting each right angle. The electric field lines are straight within each quadrant: in the first quadrant, for example, $dy/dx=-1$, so $x+y$ is constant along a line. They start on $x=0$ and finish on $y=0$. At a sheet, the normal component has the usual surface-charge jump, so the quadrant formula specifies the appropriate one-sided limits rather than a unique electric field on the charged surface.

Initially the isolated sphere has $Q=4\pi {ϵ}_{0}RV$. After insertion without contact, its charge remains $Q$. In the long-cylinder approximation, with the sphere far from the openings, neglect electric flux escaping through the ends. A Gauss surface running through the shielding metal and closed by end sections of negligible flux then encloses zero net charge. Its cavity wall therefore carries $-Q$. Since that cylinder is isolated and initially neutral, its outside wall carries $+Q$. The grounded outer cylinder supplies charge $-Q$ on its inside wall, terminating the annular electric field lines.

For positive $V$, the cavity electric field lines run from the sphere to the inner wall, while annular electric field lines run from the outside of the inner conductor to the grounded outer conductor. Negative $V$ reverses their directions. The transverse picture is schematic: exact cavity fields depend on the sphere's position and the ends. The following annular formulas neglect end effects and place the sphere well inside the long cylindrical assembly.

The electric dipole moment is $\mathbf{p}=qa\hat{\mathbf{y}}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\mathbf{\theta }}$ pointing toward increasing $\theta$. The exact electric potential isFor $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the electric dipole approximationThe radial electric field follows from $E_r=-\partial \varphi /\partial r$, and the tangential electric field from $E_{\theta }=-(1/r)\partial \varphi /\partial \theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test electric charge gives the force componentsThe electric field lines emerge from the positive electric charge and terminate on the negative electric charge. They are mirror-symmetric about the electric dipole axis and bow out across its equatorial plane. In the point-electric dipole limit their meridional curves obey $dr/(rd\theta )=E_r/E_{\theta }=2\cot \theta$, hence $r=C{\sin }^2\theta$.

In a Cartesian coordinate system the leading electric dipole field givesThereforeThe force is downwards on the equator and upwards on the negative axial side. A free positive electric charge released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the force also acquires a positive $x$ component. Its path bends toward the right; the force continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects radiation reaction.

There is a useful exact result within the point-electric dipole approximation. Put $A=k_{e}pQ>0$; the potential energy is $U=Ay/r^3$, a homogeneous function of degree $-2$. Thus $\mathbf{r}\cdot \mathbf{F}=2U$. For a particle of mass $m$, Newton's second law and conservation of energy implyAt release, both kinetic energy and potential energy vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos \alpha$, $y=-d\sin \alpha$. ThenThe electric charge gains speed until the bottom point, where the force is inward and the speed is $\sqrt{2A/(m{d}^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,Its trajectory is not an electric field line: the force determines acceleration, rather than velocity direction. For the finite two-charge electric dipole, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.

Use $x=\rho \cos \varphi$, $y=\rho \sin \varphi$ to recognize the planar quadrupole potential, $V=A(x^2-y^2)$. Its electric field is minus the gradient:or, in cylindrical coordinates,For $A>0$, arrows point toward the origin along the $x$ axis and away from it along the $y$ axis. The off-axis electric field lines satisfyThe axes themselves are the separating field lines; the origin is a field null. Reversing $A$ reverses every arrow. The field is independent of $z$, and $A$ has units $\mathrm{V}{\mathrm{m}}^{-2}$.